\newproblem{lay:5_4_26}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.4.26}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	It can be shown that the trace of a matrix equals the sum of its eigenvalues. Verify this statement for the case when $A$ is diagonalizable.
}{
  % Solution
	If $A$ is diagonalizable, then $A=PDP^{-1}$, that is $D$ is similar to $A$. Then, by Exercise 5.4.25
	\begin{center}
		$\mathrm{tr}\{A\}=\mathrm{tr}\{D\}=\sum\limits_{i=1}^n{\lambda_i}$
	\end{center}
}
\useproblem{lay:5_4_26}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
